Civil Engineering Reference
In-Depth Information
e = 3.047 × 210000 × 11.9 2 /( 2 × 275 × 573.0 ) = 287.5 > 100
EC3-1-5, 6.5(3)
y 1 = 100 + 19.6 × ( 18.54 / 2 + ( 100 / 19.6 ) 2 + 17.09 ) = 241.9
EC3-1-5, 6.5(3)
y 2 = 100 + 19.6 × ( 18.54 + 17.09 ) = 217.0 < 241.9 EC3-1-5, 6.5(3)
y = 217.0mm.
EC3-1-5, 6.5(3)
217.0 × 11.9 × 275
1694 × 10 3
λ F =
= 0.648 > 0.5
EC3-1-5, 6.4(1)
χ F = 0.5 / 0.648 = 0.772 EC3-1-5, 6.2
L eff = 0.772 × 217.0 = 167.6mm EC3-1-5, 6.2
F Rd = 275 × 167.6 × 11.9 / 1.0N = 548kN EC3-1-5, 6.2
< 558kN = R Ed , and so the assumed value of ( s s + c ) = 100 mm
is inadequate.
5.12.16 Example 16 - designing a cantilever
Problem . A3.0mlongcantileverhasthenominalimposedanddeadloads(which
include allowances for self weight) shown in Figure 5.44c. Design a suitable UB
section in S275 steel if the cantilever has sufficient bracing to prevent lateral
buckling.
Selecting a trial section .
M Ed = 1.35 × ( 40 × 3 + 20 × 3 2 / 2 ) + 1.5 × ( 120 × 3 + 10 × 3 2 / 2 )
= 891 kNm.
Assume that f y = 265 MPa and that the section is Class 2.
W pl , y M Ed / f y = 891 × 10 6 / 265mm 3 = 3362cm 3
6.2.5
Choose a 610 × 229 UB 125 with W pl , y = 3676 cm 3 > 3362 cm 3 .
Checking the trial section . The trial section must now be checked by classifying
thesectionandcheckingformoment,shear,momentandshear,andbearing.This
process is similar to that used in Section 5.12.15 for checking a simply supported
beam. If the section chosen fails any of the checks, then a new section must be
chosen.
 
Search WWH ::




Custom Search